1. Pyruvate dehydrogenase is a large comples enzyme containing many .pdf

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1. Pyruvate dehydrogenase is a large comples enzyme containing many copies of each of three enzyme called E1, E2 and E3. 2. Pyruvate dehydrogenase(E1) has TPP prosthetic groups. Dihyrolipoyl transacetylase(E2) has Lipoamide prosthetic groups, similarly Dihydrolipyl dehydrogenase(E3) has FAD prosthetic group. 3. Dihyrolipoamide and lipoamide can be interconverted. Lipoamide consists of lipoic acid linked via an amid bond to sigma amino groups of a Lys residue. 4. Hydroxyethyl group derived from pyruvate attacks the lipoamide disulfide, and TPP is eliminated. Dihydroxyethyl carbanion oxidised to acetyle group and lipoamide disulfide is reduced. 5. E2 then catalyses to transfer of acetyle group to CoA to acetyle CoA and dihydrolipoamideE2. 6. Dihydrolipoyl dehydrogenase reoxidizes dihydrolipoamide to complete catalytic cycle of E2. The oxidized E3 contains a reactive Cys-Cys disulfide bond and a tightly bound FAD. 7. Finally, reduced E3 is reoxidized. The sulfhydryl groups are reoxidized by a mechanish in which FAD funnels electrons to NAD+ yielding NADH. Solution 1. Pyruvate dehydrogenase is a large comples enzyme containing many copies of each of three enzyme called E1, E2 and E3. 2. Pyruvate dehydrogenase(E1) has TPP prosthetic groups. Dihyrolipoyl transacetylase(E2) has Lipoamide prosthetic groups, similarly Dihydrolipyl dehydrogenase(E3) has FAD prosthetic group. 3. Dihyrolipoamide and lipoamide can be interconverted. Lipoamide consists of lipoic acid linked via an amid bond to sigma amino groups of a Lys residue. 4. Hydroxyethyl group derived from pyruvate attacks the lipoamide disulfide, and TPP is eliminated. Dihydroxyethyl carbanion oxidised to acetyle group and lipoamide disulfide is reduced. 5. E2 then catalyses to transfer of acetyle group to CoA to acetyle CoA and dihydrolipoamideE2. 6. Dihydrolipoyl dehydrogenase reoxidizes dihydrolipoamide to complete catalytic cycle of E2. The oxidized E3 contains a reactive Cys-Cys disulfide bond and a tightly bound FAD. 7. Finally, reduced E3 is reoxidized. The sulfhydryl groups are reoxidized by a mechanish in which FAD funnels electrons to NAD+ yielding NADH..

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Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active Solution Yes. The compound is actively active. It is mainly caused by the carbon to the right of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices over 3 possible carbons) which give you 8. And as I previously explained: To be optically active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3 such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally active since they have 4 different groups. However, since they are mirror images of each other, they cancel each other out. It is the third and final carbon to the right of both oxygens, that makes this compound chirally active. Since it has also 4 different groups attached to the carbon and is not canceled out by anything. One simple way to test if something is optically active is to see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center, then it is optically active.

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When atoms share one or more pairs of electrons t.pdf

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This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the solution. Now the two are separate. To get the NaCl out of the water just boil it off. Solution This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the solution. Now the two are separate. To get the NaCl out of the water just boil it off..

This is quite simple to do. Barium sulfate is ins.pdf

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Step1 Cocentration of NaCl =.250 moles Step2 Mol.pdf

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